Question

A particle is performing circular motion of radius 1 m. Its speed is v = (2t^2) m/s. What will be the magnitude
of its acceleration at t = 1s ?
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Answer 1

Answer:

$4 \sqrt{2} \: m/ {s}^{2}$

Explanation:

For tangential acceleration

$= > a_{T} = \frac{dv}{dt} \\ = > a_{T} = 4t$

At t = 1s

$\boxed{a_{T} = 4m/ {s}^{2} }$

For centripetal acceleration

$= > a_{C} = \frac{ {v}^{2} }{r} \\ = > a_{C} = \frac{4 {t}^{4} }{1} \\ = > a_{C} = 4 {t}^{4}$

At t = 1s

$\boxed{a_{C} = 4m/ {s}^{2} }$

Net acceleration, (a)

$= > a = \sqrt{ {a_{T}}^{2} + {a_{C}}^{2} } \\ = > a = \sqrt{ {4}^{2} + {4}^{2} } \\ = > a = \sqrt{16 + 16} \\ = > a = \sqrt{32 } \\ = > \boxed{a = 4 \sqrt{2} \: m/ {s}^{2} }$

Answer 2

Answer:

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when particle perform circular motion it has two acceleration

1)centripetal acceleration (ac)

2)tangential acceleration( at)

$v \: = 2t {}^{2} \\ \\ a \: = 4t{}^{4} \\ \\ at \: time \: = 1sec \\ \\ a \: = 4 ms {}^{ - 2}$

Tangential acceleration :

dv/dt

= 4ms^-2

$\\ \\ \huge \bold{ \frac{v {}^{2} }{r} }$

Net acceleration :

$\huge \bold{ \sqrt{ac {}^{2} + at {}^{2} } } \\ \\ \sqrt{4 {}^{2} + 4 {}^{2} } \\ \\ \sqrt{16 + 16 } \\ \\ \sqrt{32 } \\ \\ 4 \sqrt{2} ms {}^{ - 2}$

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